16_ch 15 Mechanical Design budynas_SM_ch15

# 16_ch 15 Mechanical Design budynas_SM_ch15 - The stress is...

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394 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design (c) Eq. (15-33): C s = 1190 477 log 7 . 0 = 787 Eq. (15-36): C m = 0 . 0107 ± 56 2 + 56(56) + 5145 = 0 . 767 Eq. (15-37): C v = 0 . 659 exp[ 0 . 0011(679 . 8)] = 0 . 312 Eq. (15-38): ( W t ) all = 787(7) 0 . 8 (2)(0 . 767)(0 . 312) = 1787 lbf Since W t G < ( W t ) all , the mesh will survive at least 25 000 h. Eq. (15-61): W f = 0 . 025(966) 0 . 025 sin 4 . 764° cos 20° cos 4 . 764° =− 29 . 5 lbf Eq. (15-63): H f = 29 . 5(679 . 8) 33 000 = 0 . 608 hp H W = 106 . 4(677 . 4) 33 000 = 2 . 18 hp H G = 966(56 . 45) 33 000 = 1 . 65 hp The mesh is sufﬁcient Ans. P n = P t / cos λ = 8 / cos 4 . 764 = 8 . 028 p n = π/ 8 . 028 = 0 . 3913 in σ G = 966 0 . 3913(0 . 5)(0 . 125) = 39 500 psi
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Unformatted text preview: The stress is high. At the rated horsepower, σ G = 1 1 . 65 39 500 = 23 940 psi acceptable (d) Eq. (15-52): A min = 43 . 2(8 . 5) 1 . 7 = 1642 in 2 < 1700 in 2 Eq. (15-49): H loss = 33 000(1 − . 7563)(2 . 18) = 17 530 ft · lbf/min Assuming a fan exists on the worm shaft, Eq. (15-50): ¯ h C R = 1725 3939 + . 13 = . 568 ft · lbf/(min · in 2 · ◦ F) Eq. (15-51): t s = 70 + 17 530 . 568(1700) = 88 . 2 ◦ F Ans....
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